Question

Show that the function f : R − {3} → R − {2} given by $f\left(x\right)=\frac{x-2}{x-3}$ is a bijection.

Answer 1

f : R − {3} → R − {2} given by
$f\left(x\right)=\frac{x-2}{x-3}$
Injectivity:
Let x and y be any two elements in the domain (R − {3}), such that f(x) = f(y).
f(x) = f(y)
$$\begin{gathered} \Rightarrow \frac{x-2}{x-3}=\frac{y-2}{y-3} \\ \Rightarrow \left(x-2\right)\left(y-3\right)=\left(y-2\right)\left(x-3\right) \\ \Rightarrow xy-3x-2y+6=xy-3y-2x+6 \\ \Rightarrow x=y \end{gathered}$$
So, f is one-one.

Surjectivity:
Let y be any element in the co-domain (R − {2}), such that f(x) = y for some element x in R − {3} (domain).
f(x) = y
$$\begin{gathered} \Rightarrow \frac{x-2}{x-3}=y \\ \Rightarrow x-2=xy-3y \\ \Rightarrow xy-x=3y-2 \\ \Rightarrow x\left(y-1\right)=3y-2 \\ \Rightarrow x=\frac{3y-2}{y-1},\text{which is in}R-\left\{3\right\} \end{gathered}$$
So, for every element in the co-domain, there exists some pre-image in the domain.
$\Rightarrow$f is onto.
Since, f is both one-one and onto, it is a bijection.